A generalization of $b$-weakly compact operators

TL;DR

This paper clarifies the distinction between $b$-weakly compact and $KB$-operators, providing new insights into their properties and relationships within normed vector lattices.

Contribution

It resolves an open problem by proving that $b$-weakly compact and $KB$-operators are different classes of operators.

Findings

$KB$-operators have specific convergence properties.

$b$-weakly compact operators are a distinct class.

Relationships between $KB$-operators and $b$-weakly compact operators are characterized.

Abstract

A. Bahramnezhad and K. Haghnejad Azar introduced the classes of $KB$-operators and $WKB$-operators, and they studied some of theirs properties. In the present paper, we give answer for an open problem from that paper, which two classifications of operators, $b$-weakly compact operators and $KB$-operators are different. A continuous operator $T$ from a normed vectoe lattice $E$ into a normed space $X$ is said to be $KB$-operator (respectively, $WKB$-operator) if $\{Tx_n\}_n$ has a norm (respectively, weak) convergent subsequence in $X$ for every positive increasing sequence $\{x_n\}_n$ in the closed unit ball $B_E$ of $E$. We investigate some other properties of $KB$-operators and its relationships with $b-$weakly compact operators.

Equations

$$K(E,X)\subseteq W(E,X)\subseteq W_b(E,X)\subseteq L_{KB}(E,X)\subseteq W_{KB}(E,X).$$

$$T(f)=({\int }_0^{+\infty }f(x)\cos n\pi xdx{)}_{n=1}^{+\infty }\text{for all}f\in E.$$

$$|{\int }_0^{+\infty }f(x)\cos n\pi xdx|\le {\int }_0^{+\infty }|f(x)|dx<+\infty .$$

$$\Vert T_n-T\Vert \to 0$$

$$\begin{gathered} \Vert T{x}_{m,m}-T{x}_{k,k}\Vert \le \Vert T{x}_{m,m}-T_n{x}_{m,m}\Vert +\Vert T_n{x}_{m,m}-T_k{x}_{m,m}\Vert \\ +\Vert T_k{x}_{m,m}-T{x}_{m,m}\Vert , \end{gathered}$$

$$\{y\in G:0\le y\le x_n\}\uparrow x_n,$$

$$\Vert T{x}_n-T{z}_{mn}\Vert \le \Vert T\Vert \Vert x_n-z_{mn}\Vert \le \Vert T\Vert \Vert x_n-y_{mn}\Vert \to 0.$$

$$\Vert T{x}_n-y\Vert \le \Vert T{x}_n-T{z}_{mn}\Vert +\Vert T{z}_{mn}-y\Vert .$$

$$T\in L_{KB}(E,F)\cap W_b(E,F)$$

$$B(E,F)=L_{KB}(E,F)$$

$$\Vert x{\Vert }_{\infty }=\inf \{\lambda >0:|x|\le \lambda e\}$$

$${\sup }_n\Vert x_n{\Vert }_{\infty }={\sup }_n\Vert x_n\Vert .$$

$$0<\lambda =\text{sup}{\lambda }_n\le {\sup }_n\Vert x_n{\Vert }_{\infty }+1<{\sup }_n\Vert x_n\Vert +1<\infty .$$

$${\sup }_n\Vert x_n\Vert <+\infty .$$

$$0$$

$$\le z_{\beta }+z_{\beta }\downarrow 0,$$

$$KU(X,F)$$

$$K^{\sigma }U(X,F)$$

$$W_{KB}(E,F)\subseteq L_{UKB}(E,F).$$

Full text

1. Introduction and preliminaries

A subset $A$ of a vector lattice $E$ is called $b$-order bounded in $E$ if it is order bounded in $E^{\sim\sim}$. An operator $T:E\rightarrow X$, mapping each $b$-order bounded subset of $E$ into a relatively weakly compact subset of $X$ is called a $b$-weakly compact operator. Alpay, Altin and Tonyali introduced the class of $b$-weakly compact operators for vector lattices having separating order duals [3]. In [4], Alpay and Altin proved that a continuous operator $T$ from a Banach lattice $E$ into a Banach space $X$ is $b$-weakly compact if and only if $\{Tx_{n}\}_{n}$ is norm convergent for each $b$-order bounded increasing sequence $\{x_{n}\}_{n}$ in $E^{+}$ if and only if $\{Tx_{n}\}_{n}$ is norm convergent to zero for each $b$-order bounded disjoint sequence $\{x_{n}\}_{n}$ in $E^{+}$. Authors in [6] proved that an operator $T$ from a Banach lattice $E$ into a Banach space $X$ is $b$-weakly compact if and only if $\{Tx_{n}\}_{n}$ is norm convergent for every positive increasing sequence $\{x_{n}\}_{n}$ of the closed unit ball $B_{E}$ of $E$. The aim of this paper is studied the classes of operators on Banach lattices are called $KB$-operators and $WKB$-operators. We will investigate on theirs properties. A continuous operator $T$ from a Banach lattice $E$ into a normed space $X$ is said to be $KB$-operator ( $WKB$-operato), if $\{Tx_{n}\}_{n}$ has a norm ( weak) convergent subsequence in $X$ for every positive increasing sequence $\{x_{n}\}_{n}$ in the closed unit ball $B_{E}$ of $E$, see [9]. To state our results, we need to fix some notation and recall some definitions.

A Banach lattice $E$ is is said to be an $AM$-space if for each $x,y\in E$ such that $|x|\wedge|y|=0$, we have $\|x+y\|=max\{\|x\|,\|y\|\}$. A Banach lattice $E$ is an $AL$-space if its topological dual $E^{\prime}$ is an $AM$-space. A Banach lattice $E$ is said to be $KB$-space whenever each increasing norm bounded sequence of $E^{+}$ is norm convergent. An operator $T:E\rightarrow F$ between two Riesz spaces is positive if $T(x)\geq 0$ in $F$ whenever $x\geq 0$ in $E$. Note that each positive linear mapping on a Banach lattice is continuous. An operator $T$ from a Banach space $X$ into a Banach space $Y$ is compact (resp. weakly compact) if $\overline{{T(B_{X})}}$ is compact (resp. weakly compact) where $B_{X}$ is the closed unit ball of $X$. For terminology concerning Banach lattice theory and positive operators, we refer the reader to the excellent book of [1].

As $b$-weakly compact operators [7], the class of $KB$-operators and $WKB$-operators does not satisfy duality property. In fact the identity operator of the Banach lattice $\ell^{1}$ is a $KB$-operator (respectively, $WKB$-operator); but its adjoint which is the identity operator of the Banach lattice $\ell^{\infty}$, is not a $KB$-operator (respectively, $WKB$-operator). Conversely, the identity operator of the Banach lattice $c_{0}$ is not a $KB$-operator (respectively, not $WKB$-operator); but its adjoint, which is the identity operator of the Banach lattice $\ell^{1}$, is a $KB$-operator (respectively, $WKB$-operator). Recall that a Banach space is said to have Schur property whenever every weak convergent sequence is norm convergent, i.e., whenever $x_{n}\xrightarrow{w}0$ implies $\|x_{n}\|\rightarrow 0$. Let $E$, $F$ be Banach lattices. If either $E$ or $F$ has the Schur property then $L(E,F)=W_{b}(E,F)$ [4]. It is also clear that for a Banach lattice $E$ and a Banach space $X$ with Schur property, every $WKB$-operator $T:E\rightarrow X$ is a $KB$-operator. Let $E$ be a vector lattice. A sequence $\{x_{n}\}_{1}^{\infty}\subset E$ is called order convergent to $x$ as $n\to\infty$ if there exists a sequence $\{y_{n}\}^{\infty}_{1}$ such that $y_{n}\downarrow 0$ as $n\rightarrow\infty$ and $|x_{n}-x|\leq y_{n}$ for all $n$. We will write $x_{n}\xrightarrow{o_{1}}x$ when $\{x_{n}\}$ is order convergent to $x$. A sequence $\{x_{n}\}$ in a vector lattice $E$ is strongly order convergent to $x\in E$, denoted by $x_{n}\xrightarrow{o_{2}}x$ whenever there exists a net $\{y_{\beta}\}_{\beta\in\mathcal{B}}$ in $E$ such that $y_{\beta}\downarrow 0$ and that for every $\beta\in\mathcal{B}$, there exists $n_{0}$ such that $|x_{n}-x|\leq y_{\beta}$ for all $n\geq n_{0}$. It is clear that every order convergent sequence is strongly order convergent, but two convergence are different, for information see, [2]. A net $(x_{\alpha})_{\alpha}$ in Banach lattice $E$ is unbounded norm convergent (or, $un$-convergent for short) to $x\in E$ if $|x_{\alpha}-x|\wedge u\xrightarrow{\|.\|}0$ for all $u\in E^{+}$. We denote this convergence by $x_{\alpha}\xrightarrow{un}x$. This convergence has been introduced and studied in [10, 13].

2. Main Results

In each parts of this manuscript, $E$ is a normed vector lattice and $X$ normed space. The definition of $b-$weakly compact operator holds whenever $E$ is a normed vector lattice and $X$ normed space. The collections of $KB$-operators, $WKB$-operators, b-weakly compact operators, order weakly compact operators, weakly compact operators and compact operators will be denoted by $L_{KB}(E,X)$, $W_{KB}(E,X)$, $W_{b}(E,X)$, $W_{o}(E,X)$, $W(E,X)$ and $K(E,X)$, respectively, whenever there is not confused. By notice an example from [12], page 95, the classification of order compact or order weakly compact operators from Banach lattice $E$ into Banach space $X$, in general, are not subspace of $L_{KB}(E,X)$, but by Theorem 3.4.4, from [12], every interval-bounded and order weakly compact operator $T:E\rightarrow X$ is $b-$weakly compact operator, and so $KB-$operator. On the other hand, if $e\in E$ is an atom, then $L_{KB}(E_{e},F)=W_{o}(E_{e},F)$. We have the following relationships between these spaces:

[TABLE]

Let $E$ be normed vector lattices, $X$ normed space and $T:E\rightarrow X$ be a positive operator. By using Prpostion 2.1 from [9], $T\in W_{b}(E,X)$ iff $T\in L_{KB}(E,X)$ iff $T\in W_{KB}(E,X)$ as follows.

Proposition 2.1**.**

[9]** Let $E$ and $F$ be Banach lattices and $T:E\rightarrow F$ be a positive operator. Then the following statements are equivalent.

1. (1)

$T$* is $b$-weakly compact.* 2. (2)

$T$* is $KB$-operator.* 3. (3)

$T$* is $WKB$-operator.*

In the following example, we give answer for an open problem from [9], which two classifications of operators $W_{b}(E,X)$ and $L_{KB}(E,X)$ are different. In the other words, there is a $KB$-operator $T$ from normed vector lattice $E$ into Banach space $X$ that is not $b-$weakly compact operator.

Example 2.2**.**

In the following, we show that the inclusion $W_{b}(E,F)\subseteq L_{KB}(E,F)$ may be proper whenever $E$ is normed vector lattice and $F$ is a Banach space.

Proof.

Authors in Theorem 2.10 of [6] proved that an operator $T$ from Banach lattice $E$ into Banach lattice $F$ is $b-$weakly compact if and only if $(T(x_{n}))_{n=1}^{+\infty}$ is norm convergent for every positive increasing sequence $(x_{n})$ in the closed unit ball $B_{E}$ of $E$. Now this theorem holds when $E$ is normed vectoe lattice.

Suppose that $E=\langle\{\chi_{[0,r]}\ :\ r\in\mathbb{R}^{+}\}\rangle$ where $\chi_{[0,r]}$ is a characteristic function on $[0,r]$. It is clear that $E$ is a subspace of $L^{\infty}([0,+\infty))$ and with $\|.\|_{\infty}$ is normed vector lattice. We define an operator $T:E\rightarrow l^{\infty}$ with

[TABLE]

$T$ is a bounded linear operator of course

[TABLE]

Now the proof will be based upon three claims as follows:

claim 1: $T$ is not positive operator.

If we set $f=\chi_{[0,\frac{1}{2}]}$, then $Tf=(\frac{1}{n\pi}\sin\frac{n\pi}{2})_{n=1}^{+\infty}$ which shows that $T$ is not positive operator.

claim 2: $T$ is not $b-$weakly compact operator.

Let $f_{m}=\chi_{[0,m+\frac{1}{2}]}$ where $m\in\mathbb{N}$. Then $0\leq f_{m}\uparrow$ and $\mathop{\text{sup}}_{m}\|f_{m}\|_{\infty}<+\infty$, but $(Tf_{m})_{m=1}^{+\infty}$ is not convergent.

claim 3: $T$ is a $KB-$operator.

Let $\{f_{m}\}\subseteq E$, $0\leq f_{m}\uparrow$ and $\mathop{\text{sup}}_{m}\|f_{m}\|_{\infty}<+\infty$. We write $Tf_{m}=a_{m,n}$ where $a_{m,n}=\int_{0}^{+\infty}f_{m}(x)\cos nxdx$.

For fix $n\in\mathbb{N}$, the sequence $(a_{m,n})_{m=1}^{+\infty}$ is norm bounded in $\mathbb{R}$, and so it has a subsequence $(a_{k_{m},n})_{m=1}^{+\infty}$ which is convergent. It is clear $0\leq f_{k_{m}}\uparrow$ and $\mathop{\text{sup}}_{m}\|f_{k_{m}}\|_{\infty}<+\infty$. It follows that $(a_{k_{m},n+1})_{m=1}^{+\infty}$ is norm bounded in $\mathbb{R}$, and so it has also subsequence $(a_{(k+1)_{m},n+1})_{m=1}^{+\infty}$ which is convergent. Thus $(a_{m_{m},n})_{m=1}^{+\infty}$ is a subsequence of $(a_{m,n})_{m=1}^{+\infty}$ which is convergent for each $n\in\mathbb{N}$. On the other hand, $(Tf_{m_{m}})_{m=1}^{+\infty}$ is subsequence of $(Tf_{m})_{m=1}^{+\infty}$ which is convergent. Now, $T$ is a $KB-$operator. ∎

Proposition 2.1, shows that $L_{KB}(E,F)=W_{b}(E,F)$ whenever $L_{KB}(E,F)$ is vector lattice. Consequently, the above example implies that in general $L_{KB}(E,F)$ is not vector lattice. On the other hand, in general, $L_{KB}(E,F)$ is not order dense in $B(E,F)$. Note that if we set $E=F=c_{0}$ and $T_{n}\in L_{KB}(c_{0},c_{0})$ defined by $T_{n}(x_{1},x_{2},...)=(x_{1},x_{2},...,x_{n},0,0,...)$, then $0\leq T_{n}\uparrow I_{c_{0}}$. But $I_{c_{0}}$ is not $KB$-operator. Then by Proposition 2.1, $W_{b}(E,X)$, in general, is also not order dense in $B(E,F)$.

In the following we show that $L_{KB}(E,F)$ is a norm closed subspace of $B(E,F)$ whenever $F$ is a Banach lattice.

Theorem 2.3**.**

The collection of all $KB$-operators from $E$ into Banach lattice $F$ is norm closed vector subspace of $B(E,F)$.

Proof.

Let $\{T_{n}\}\subseteq B(E,F)$ be a sequence of $KB$-operators such that

[TABLE]

holds in $B(E,F)$ and let $\{x_{m}\}$ be a positive increasing sequence in $E$ with $\text{sup}_{m}\|x_{m}\|<\infty$. Since $T_{1}\in L_{KB}(E,F)$, there is a subsequence $\{x_{m,1}\}$ of $\{x_{m}\}$ which $\{T_{1}x_{m,1}\}$ is norm convergent in $F$. Choose subsequence $\{x_{m,n}\}$ and $\{x_{m,{n+1}}\}$ of $\{x_{m}\}$ as follows

1. (1)

$\{x_{m,{n+1}}\}$ is a subsequence of $\{x_{m,n}\}$ for each $n\in\mathbb{N}$.. 2. (2)

$\{Tx_{m,n}\}_{m}$ is norm convergent in $F$ for each $n\in\mathbb{N}$.

It follows that $\{T_{n}x_{m,m}\}_{m}$ is norm convergent for each $n\in\mathbb{N}$. From following inequalities

[TABLE]

the sequence $\{Tx_{m,m}\}_{m}$ is a norm Cauchy, and so norm convergent in $F$. ∎

Theorem 2.4**.**

Let $E$ and $F$ be Banach lattices where $E$ has order continuous norm. Let $G$ be a sublattice order dense in $E$ and $T$ be a positive operator from $E$ into $F$. If $T\in L_{KB}(G,F)$, then $T\in L_{KB}(E,F)$.

Proof.

By using Propostion 2.1 from [9], it is enough to give a proof for $b-$weakly compact operators. Let $\{x_{n}\}$ be a positive increasing sequence in $E$ with $\text{sup}_{n}\|x_{n}\|<\infty$. Since $G$ is order dense in $E$, by Theorem 1.34 from [1], we have

[TABLE]

for each $n$. Let $\{y_{mn}\}_{m=1}^{\infty}\subset G$ with $0\leq y_{mn}\uparrow x_{n}$ for each $n$. Put $z_{mn}=\vee_{i=1}^{n}y_{mi}$ and $0\leq T\in L(E,F)$. Its follows that $z_{mn}\uparrow_{m}x_{n}$ and $\text{sup}_{m,n}\|z_{m,n}\|\leq\text{sup}_{n}\|x_{n}\|<\infty$. Now, if $T\in W_{b}(G,F)$, then $\{Tz_{mn}\}$ is norm convergent to some point $y\in F$. Now, we have the following inequalities

[TABLE]

Thus by the following inequality proof holds

[TABLE]

∎

By notice to Remark 2.26 from [9], $L_{KB}(E,F)$ is not vector lattice whenever $E$ and $F$ are Banach lattices. On the other hand, if $T\in L_{KB}(E,F)$, in general, the modulas of $T$ need not be a $KB-$operators. Now in the following theorem, we show that $T\in L_{KB}(E,F)$ whenever $|T|\in L_{KB}(E,F)$.

Theorem 2.5**.**

Let $E$ and $F$ be normed vector lattices. We have the following assertions.

1. (1)

*If $T:E\rightarrow F$ is an order bounded operator and $F$ is *$KB-$*space, then $T$ and $|T|$ are *$KB-$operators. 2. (2)

*If $|T|$ is *$KB-$*operator or *$b-$weakly compact operator, then

[TABLE]

Proof.

1. (1)

Let $x_{n}\uparrow$ and $\mathop{\text{sup}}_{n}\|x_{n}\|<\infty$. Since $T^{+}$ is positive, by Theorem 4.3 [1], $T^{+}$ is norm bounded. It follows that $\{T^{+}x_{n}\}$ is norm bounded. Thus $T^{+}$ is $KB-$operator. In similar way, $T^{-}$ is also $KB$-operator. It follows that $T$ and $|T|$ are $KB-$operators. 2. (2)

Since $0\leq T^{-},T^{+}\leq|T|$, then $T^{-}$ and $T^{+}$ are $KB-$operators. By Proposition 2.9, from [9], $T$ is $KB-$operators. Similar argument holds for $b-$weakly compact operators, and so by using proposition 2.1, proof holds.

∎

Theorem 2.5 shows that the modulus of the operator $T$ in example of page 3 is not $KB-$operator or $b-$weakly compact operator.

Note that each weakly compact operator is a $KB$-operator but the converse may be false in general. For example, the identity operator $I:L^{1}[0,1]\rightarrow L^{1}[0,1]$ is a $KB$-operator but is not weakly compact.

Let $E$ and $F$ be two Banach lattices such that the norm of $E^{\prime}$ is order continuous. Then, by using Proposition 2.1 and [8, Theorem 2.3], it is clear that each positive $KB$-operator $T:E\rightarrow F$ is weakly compact. Now in the following, we show that $B(E,F)=L_{KB}(E,F)$ whenever $E$ has order unit and order continuous norm.

Theorem 2.6**.**

Let $E$ be a normed vectoe lattice and $F$ Dedekind Banach Lattice. By one of the following conditions

[TABLE]

1. (1)

$E$* has order unit and order continuous norm.* 2. (2)

$E$* and $E^{\prime}$ have order continuous norm and $F$ has Schur property*

Proof.

1. (1)

Let $T\in B(E,F)$ and $x_{n}\uparrow$ and $\mathop{\text{sup}}_{n}\|x_{n}\|<\infty$. Let $e\in E^{+}$ be an order unit for $E$. For each $x\in E$, the norm

[TABLE]

is equivalent to the original norm $\|.\|$ of $E$. It follows that

[TABLE]

For each $n\in\mathbb{N}$, there exists $\lambda_{n}>0$ such that $\lambda_{n}\leq\|x_{n}\|_{\infty}+1$ where $|x_{n}|\leq\lambda_{n}e$. It follows that

[TABLE]

Then $\{x_{n}\}\subseteq[-\lambda e,\lambda e]$. Since $E$ has order continuous norm, by using theorem 4.9 from [1], it follows that $[-\lambda e,\lambda e]$ is weakly compact. On the other hand, since $E$ has order unit, $T$ is order bounded, and so $T^{+}$ exists. Thus $T^{+}([-\lambda e,\lambda e])$ is weakly compact, since $T^{+}$ is weak to weak continuous. As $\{T^{+}x_{n}\}\subseteq T^{+}([-\lambda e,\lambda e])$, there is subsequence $\{T^{+}x_{n_{j}}\}$ from $\{T^{+}x_{n}\}$ which is weak convergent to some point $z\in F$. Since $(T^{+}x_{n})_{n}$ is an increasing sequence, then by Theorem 1.4.1 from [12], $T^{+}x_{n}$ is norm convergent to $z$ in $F$. Thus $T^{+}\in L_{KB}(E,F)$. By using equality $T=T^{+}-T^{-}$, we conclude that $T\in L_{KB}(E,F)$. 2. (2)

Let $T\in B(E,F)$ and $(x_{n})\subseteq E$ be increasing positive sequence where

[TABLE]

By Theorem 4.25 from [1], there is a weakly Cauchy subsequence $\{x_{n_{j}}\}$ in $E$. It follows that $\{Tx_{n_{j}}\}$ is weakly Cauchy in $F$. Since $F$ has Schur property, $\{Tx_{n_{j}}\}$ is norm Cauchy in $F$, and so convergence in $F$. Thus $T\in L_{KB}(E,F)$.

∎

Theorem 2.7**.**

Assume that an order bounded operator $T:E\rightarrow F$ between two Banach lattices has preserves disjointness. Then $T\in W_{b}(E,F)$ if and only if $T\in L_{KB}(E,F)$.

Proof.

By using Theorem 2.40 from [1], $T$ is $KB-$operator if and only if $|T|$ is $KB-$operator, and by proposition 2.1 from [9] it is equivalent to that $|T|$ is $b-$weakly compact operator. Another using Theorem 2.40 [1] shows that $T$ is $b-$weakly compact and proof follows. ∎

Let $E$ and $F$ be Banach lattices and let $T\in L_{KB}(E,F)$. By Proposition 3.4 from [10], $\{Tx_{n}\}_{n}$ has an order convergence subsequence in $F$ for every positive increasing sequence $\{x_{n}\}_{n}$ in the closed unit ball $B_{E}$ of $E$. Now, in the following we introduce two classifications of operators which are generalization of order and strongly order continuous operators and we establish the relationships between them and positive $KB-$operators.

Definition 2.8**.**

Let $E$ and $F$ be vector lattice. $L^{(1)}_{c}(E,F)$ (resp. $L^{(2)}_{c}(E,F)$) is the collection of operators $T\in L_{b}(E,F)$, which $x_{n}\xrightarrow{o_{1}}0$ ($x_{n}\xrightarrow{o_{2}}0$) implies $Tx_{n_{k}}\xrightarrow{o_{1}}0$ (resp. $Tx_{n_{k}}\xrightarrow{o_{2}}0$) whenever $\{x_{n_{k}}\}$ is a subsequence of $\{x_{n}\}$.

In [2], there are some examples which shows that two classifications of operators $L^{(1)}_{c}(E,F)$ and $L^{(2)}_{c}(E,F)$ are different.

Theorem 2.9**.**

Let $E$, $F$ be a Banach lattices and $E$ has order continuous norm. Then

1. (1)

$W_{b}(E,F)^{+}=L_{KB}(E,F)^{+}\subseteq L^{(2)}_{c}(E,F)$. 2. (2)

If $L_{KB}(E,F)$ is vector lattice and $F$ Dedekind complete, then $L_{KB}(E,F)$ is an ideal in $L^{(1)}_{c}(E,F)=L^{(2)}_{c}(E,F)$.

Proof.

1. (1)

Let $T$ be a positive $KB-$operator and $\{x_{n}\}\subset E$ strongly order convergence in $E$. Without lose generality, we set $0\leq x_{n}\xrightarrow{o_{2}}0$, which follows $\{x_{n}\}$ is norm convergent to zero. Set $\{x_{n_{j}}\}$ as subsequence with $\sum_{k=1}^{+\infty}\|x_{n_{j}}\|<+\infty$. Define $y_{m}=\sum_{j=1}^{m}x_{n_{j}}$. Then $0\leq y_{m}\uparrow$ and $\text{sup}_{m}\|y_{m}\|<\infty$. Since $T$ is $KB-$operator, $\{Ty_{m}\}$ is norm convergent to some point $z\in F$. Now by [11], page 7, it has a subsequence as $\{Ty_{m_{k}}\}$ which is strongly order convergent to $z\in F$. Thus there is $\{z_{\beta}\}\subset F^{+}$ and that for each $\beta$ there exists $n_{0}$ $|Ty_{m_{k}}-z|\leq z_{\beta}\downarrow 0$ whenever $k\geq n_{0}$. If we set $n_{0}\leq k^{\prime}\leq k$, then we the following inequalities

[TABLE]

shows that $T\in L^{(2)}_{c}(E,F)$ and proof immediately follows. 2. (2)

Since $L_{KB}(E,F)$ is a vector lattice, by equality $T=T^{+}-T^{-}$ and Theorem 1.7 from [2], we have $L^{(2)}_{c}(E,F)=L^{(1)}_{c}(E,F)$ and by part (1), $L_{KB}(E,F)$ is a subspace of $L^{(1)}_{c}(E,F)$. By Proposition 2.9, from [9], it is also obvious that $L_{KB}(E,F)$ is an ideal in $L^{(1)}_{c}(E,F)$.

∎

Question. By conditions Theorem 2.9, Is $L_{KB}(E,F)$ a band in $L^{(1)}_{c}(E,F)=L^{(2)}_{c}(E,F)$?

Definition 2.10**.**

1. (1)

An operator $T:E\rightarrow F$ between two normed vector lattice is unbounded $b$-weakly compact if $\{Tx_{n}\}$ is $un-$convergent for every positive increasing sequence $\{x_{n}\}_{n}$ in the closed unit ball $B_{E}$ of $E$. 2. (2)

An operator $T:E\rightarrow F$ between two normed vectoe lattice is unbounded $KB$-operator if $\{Tx_{n}\}$ is $un-$convergent for every positive increasing sequence $\{x_{n}\}_{n}$ in the closed unit ball $B_{E}$ of $E$.

For normed vector lattices $E$ and $F$, the collection of unbounded

$KB-$operators (resp. $b-$weakly compact operators) will be denoted by

$L_{UKB}(E,F)$ (resp. $UW_{b}(E,F)$).

As example of page 3, two classifications of the above operators are different. If a Banach lattice $F$ has strong unit, by using Theorem 2.3 [13], we have $L_{KB}(E,F)=L_{UKB}(E,F)$ and $W_{b}(E,F)=UW_{b}(E,F)$.

It is clear that every $KB-$operators (or $b-$weakly compact operators) are unbounded $KB-$operators (or $b-$weakly compact operators), but the following example shows that the converse in general, not holds.

Example 2.11**.**

Let $I_{c_{0}}$ be an identity mapping from $c_{0}$ into itself. Then $I_{c_{0}}$ is an unbounded $KB$-operator and unbounded $b$-weakly compact operator. But $I_{c_{0}}$ is not $KB$-operator or $b$-weakly compact operator.

Let $X$ be a Banach space, $F$ a Banach lattice, and $T\in L(X,F)$. We say that $T$ is (sequentially) $un-$compact, if for every bounded net $\{x_{\alpha}\}$ (resp. $\{x_{n}\}$ its image has a subnet (resp. subsequence) which is $un-$convergent. The collections of $un-$compact (resp. sequentially $un-$compact) will be denoted by $KU(X,F)$ (resp. $K^{\sigma}U(X,F)$), that is

[TABLE]

It is clear that $K^{\sigma}U(E,F)\subseteq L_{UKB}(E,F)$ where $E$ and $F$ are Banach lattices. As example of page 3, this inclusion may be proper.

Theorem 2.12**.**

Let $I$ be an ideal in Banach lattice $E$ and $T\in B(E,F)$, where $F$ is Banach lattice. If $T\in L_{KB}(I,F)$ is surjective homomorphism, then $T\in L_{UKB}(E,F)$.

Proof.

Let $\{x_{n}\}$ be a positive increasing sequence in $E$ with $\text{sup}_{n}\|x_{n}\|<\infty$ and let $x\in I$. Then $x_{n}\wedge x\in I$, $0\leq x_{n}\wedge x\uparrow$ and $\text{sup}_{n}\|x_{n}\wedge x\|\leq\|x_{n}\|<\infty$. Since $T\in L_{KB}(I,F)$, there is a subsequence $\{x_{n_{j}}\}$ which $\{T(x_{n_{j}}\wedge x)\}$ is convergent for each $x\in I$. As $T$ is homomorphism and surjective, $\{T(x_{n_{j}})\wedge y\}$ is convergent for all $y\in F$ and proof follows. ∎

Theorem 2.13**.**

Let $E$ be a $KB-$space and $T\in B(E,F)$ be an surjective homomorphism, where $F$ is Banach lattice. Then $T\in L_{UKB}(E,F)$

Proof.

Let $\{x_{n}\}\subset E^{+}$ be increasing sequence. Since for each $x\in E^{+}$ $\text{sup}\|x_{n}\wedge x\|<\infty$, it follows $\{x_{n}\wedge x\}$ is norm convergent, and so $\{T(x_{n}\wedge x)\}$ is norm convergent for each $x\in E^{+}$. As $T$ is surjective homomorphism, proof follows. ∎

Theorem 2.14**.**

Let $E$ has order continuous norm and atomic and $F$ normed lattice. Then

[TABLE]

Proof.

Let $\{x_{n}\}\subset E^{+}$ be increasing sequence and $\text{sup}\|x_{n}\|<\infty$. If $T\in W_{KB}(E,F)$, there is a subsequence $\{x_{n_{j}}\}$ which $\{Tx_{n_{j}}\}$ is weak convergent to some point $y\in F$. By Proposition 6.2, from [10], $\{Tx_{n_{j}}\}$ is unbounded norm convergent to $y$. Thus $T\in L_{UKB}(E,F)$. ∎

Theorem 2.15**.**

Assume that $E$ and $F$ are Banach lattices and $F$ has order continuous norm. Then

1. (1)

$W_{KB}(E,F)^{+}\subseteq L_{UKB}(E,F)$. 2. (2)

If $T:E\rightarrow F$ is surjective lattice homomorphism, then $T\in L_{UKB}(E,F)$.

Proof.

Let $\{x_{n}\}\subset E^{+}$ be increasing sequence and $\text{sup}\|x_{n}\|<\infty$. Then

1. (1)

if $T\in W_{KB}(E,F)^{+}$, $\{Tx_{n}\}$ is weakly convergence in $F$. By Proposition 6.3 from [10], $\{Tx_{n}\}$ is unbounded norm convergence in $F$, and so $T\in L_{UKB}(E,F)$. 2. (2)

let $x\in E^{+}$. Set $y_{n}=x_{n}\wedge x$, which follows that $y_{n}\uparrow\leq x$ and $\text{sup}_{n}\|y_{n}\|\leq\|x\|$. Since $T$ is lattice homomorphism, $T$ is positive, which follows $Ty_{n}\uparrow\leq Tx$. By using Theorem 4.11 [1], $\{Ty_{n}\}$ norm Cauchy, and so norm convergence in $F$. On the other hand, $T(x_{n}\wedge x)=Tx_{n}\wedge Tx$ is norm convergent and proof follows.

∎

In the preceeding theorem, if $W_{KB}(E,F)$ is a vector lattice, we conclude that $W_{KB}(E,F)$ is a subspace of $L_{UKB}(E,F)$ whenever $E$ and $F$ are Banach lattices. On the other hand, if $F^{\prime}$ has order continuous norm, then by using Theorem 6.4 [10], $L_{UKB}(E,F)^{+}\subseteq W_{KB}(E,F)$ whenever $E$ and $F$ are Banach lattices.